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2 years ago

15

I'LL MARK BRAINLIST

1 answer:

OlgaM077 [116]2 years ago

8 0

The answer to your question is 210 combinations are possible :)

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The table below shows the number of students who attend various after-school activities. After-School Activities Activity Number

KatRina [158]

Answer:

C. 95:19

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the slope of the line of (1,0) and (-1,-3)

Alinara [238K]

Slope formula: m = $\frac{(y2-y1)}{(x2-x1)}$ (Knowing that m represents the slope)

Substitute (1,0) for (x1,y1), and (-1,-3) for (x2,y2)

Slope of the line of (1,0) and (-1,-3) is:

m = $\frac{(-3-0)}{(-1-1)}$ = $\frac{-3}{-2}$ = $\frac{3}{2}$

(Simplify)

Slope of the line of (1,0) and (-1,-3) is $\frac{3}{2}$

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3 years ago

Which statement about decision making analysis is false

navik [9.2K]

Answer: Id personally say D

Step-by-step explanation:

6 0

2 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

Graph a line with a slope of 3/4 that contains the point (2,-3).

arsen [322]

Answer:

Step-by-step explanation:

First, plot the point (2, -3) on the graph. Then, use the slope to pick another point. The slope is rise over run. For your slope, the line will go 3 places up and then 4 places to the right. Using a straight-edge, follow the points and you will get the graph of the line.

6 0

3 years ago